x^4 - y^4 + z^4 + 2x^2z^2 + 3x^2 + 4z^2 + 1 = 0 <=> x^4 + 2x^2(z^2 +2) + (z^2 +2)^2 = y^4 +3 +x^2 <=> (x^2 +z^2+2)^2 = y^4 +3 +x^2 <=> (x^2 +z^2 +2 -y^2)(x^2 +z^2 +2 +y^2) = x^2 +3. do )(x^2 +z^2 +2 +y^2) ; x^2 +3 >0 nen : x^2 + 3 chia het cho x^2 +z^2 +2 + y^2 nen x^2 +3 >= x^2 +z^2 +2 + y^2 <=> z^2 + y^2 <= 1 => z^2 + y^2 = 1 hoac 0 => (z^2;y^2) = {(1;0) ; (0;1)} => (z^2;y^2) = {(1;0) ; (0;1); (0;0)} the lan luot vao roi suy ra x
$x^4-y^4+z^4+2x^2z^2+3x^2+4z^2+1=0\Leftrightarrow x^4+2x^2(z^2+2) + (z^2 +2)^2 = y^4 +3 +x^2$ $\Leftrightarrow (x^2 +z^2+2)^2 = y^4 +3 +x^2$$\Leftrightarrow (x^2 +z^2 +2 -y^2)(x^2 +z^2 +2 +y^2) = x^2 +3$. Vì $(x^2 +z^2 +2 +y^2) ; x^2 +3 >0$ nên : $x^2 + 3$ chia hết cho $x^2 +z^2 +2 + y^2$ $\Rightarrow x^2 +3\geq x^2 +z^2 +2 + y^2 $$\Leftrightarrow z^2 + y^2 <= 1 $$\Rightarrow z^2 + y^2 = 1$ or $0 $$\Rightarrow (z^2;y^2) = {(1;0) ; (0;1)}$ $\Rightarrow (z^2;y^2) = {(1;0) ; (0;1); (0;0)} $the lan luot vao roi suy ra x